Bài 1:
a) \(A + B = (7{a^2} - 4ab - {b^2}) + (2{a^2} - ab + {b^2})\)
\(\eqalign{ & = 7{{\rm{a}}^2} - 4{\rm{a}}b - {b^2} + 2{{\rm{a}}^2} - ab + {b^2} \cr & = 9{{\rm{a}}^2} - 5{\rm{a}}b. \cr} \)
b) \(A - B = (7{a^2} - 4ab - {b^2}) - (2{a^2} - ab + {b^2})\)
\(\eqalign{ & = 7{{\rm{a}}^2} - 4{\rm{a}}b - {b^2} - 2{{\rm{a}}^2} + ab - {b^2} \cr & = 5{{\rm{a}}^2} - 3{\rm{a}}b - 2{b^2}. \cr} \)
Bài 2:
a) \(5{\rm{x}} + 3(3{\rm{x}} + 7) - 35 \)\(\;= 5{\rm{x}} + 9{\rm{x}} + 21 - 35 = 14{\rm{x}} - 14.\)
Ta có \(14{\rm{x}} - 14 = 0 \Rightarrow 14{\rm{x}} = 14 \Rightarrow x = 1.\)
b) \({x^2} + 8x - ({x^2} + 7x + 8) - 9 \)\(\;= {x^2} + 8x - {x^2} - 7x - 8 = x - 17\).
Ta có \(x - 17 = 0 \Rightarrow x = 17.\)
Bài 3: \(x = - 1\) là nghiệm của đa thức
\(\eqalign{ & \Rightarrow P( - 1) = 0 \Rightarrow {( - 1)^2} + 2.( - 1) + m - 1 = 0 \cr & \Rightarrow 1 - 2 + m - 1 = 0 \Rightarrow m = 2. \cr} \)
Bài 4:
\(2({x^2} - 2{\rm{x}}y) - M = 6{{\rm{x}}^2} + 5{\rm{x}}y - {y^2}\)
\(\eqalign{ & \Rightarrow M = (2{x^2} - 4{\rm{x}}y) - 6{{\rm{x}}^2} - 5{\rm{x}}y + {y^2} \cr & \Rightarrow M = 2{x^2} - 4{\rm{x}}y - 6{{\rm{x}}^2} - 5{\rm{x}}y + {y^2} \cr & \Rightarrow M = - 4{{\rm{x}}^2} - 9{\rm{x}}y + {y^2}. \cr} \)
Bài 5: Ta có:
\(\eqalign{ & f(x) = g(x)\cr& \Rightarrow {x^3} + 4{{\rm{x}}^2} - 3{\rm{x}} + 2 = {x^2}(x + 4) + x - 5 \cr & \Rightarrow {x^3} + 4{{\rm{x}}^2} - 3{\rm{x}} + 2 = {x^3} + 4{x^2} + x - 5 \cr & \Rightarrow {x^3} + 4{{\rm{x}}^2} - 3{\rm{x}} + 2 - {x^3} - 4{x^2} - x + 5 = 0 \cr & \Rightarrow - 4x = - 7 \Rightarrow x = {7 \over 4}. \cr} \)
